Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10010;
void kill() {
  puts("-1");
  exit(0);
}
int N;
int lsb(int x) { return x & -x; }
bool comp(int x, int y) { return lsb(x) < lsb(y); }
struct data {
  int arr[MAXN];
  vector<pair<int, int> > basis;
  vector<pair<int, int> > ops;
  void act(int x, int y) {
    arr[x] ^= arr[y];
    ops.push_back(pair<int, int>(x, y));
  }
  bool add(int pos) {
    int x = arr[pos];
    for (pair<int, int> pv : basis) {
      int v = pv.first;
      if (x & lsb(v)) {
        x ^= v;
        act(pos, pv.second);
      }
    }
    if (x == 0) {
      return false;
    }
    for (pair<int, int> &pv : basis) {
      int &v = pv.first;
      if (v & lsb(x)) {
        v ^= x;
        act(pv.second, pos);
      }
    }
    assert(arr[pos] == x);
    basis.emplace_back(x, pos);
    return true;
  }
  void swap(int x, int y) {
    if (x != y) {
      act(x, y);
      act(y, x);
      act(x, y);
    }
  }
  void setv(int x, int y) {
    act(x, x);
    act(x, y);
  }
  void compute() {
    vector<int> vpos;
    for (int i = 0; i < N; i++) {
      scanf("%d", &arr[i]);
      if (add(i)) {
        vpos.push_back(i);
      }
    }
    sort((basis).begin(), (basis).end(),
         [](pair<int, int> p1, pair<int, int> p2) {
           return comp(p1.first, p2.first);
         });
    for (int i = 0; i < vpos.size(); i++) {
      int mnval = arr[vpos[i]], mnpos = vpos[i];
      for (int j = i + 1; j < vpos.size(); j++) {
        if (lsb(mnval) > lsb(arr[vpos[j]])) {
          mnval = arr[vpos[j]];
          mnpos = vpos[j];
        }
      }
      swap(vpos[i], mnpos);
      swap(i, vpos[i]);
    }
    for (int i = vpos.size(); i < N; i++) {
      act(i, i);
    }
  }
};
data A, B;
void transfer() {
  vector<pair<int, int> > av = A.basis, bv = B.basis;
  int asz = av.size(), bsz = bv.size();
  for (int i = 0, j = 0; j < bsz; j++) {
    while (i < asz && lsb(av[i].first) != lsb(bv[j].first)) {
      i++;
    }
    if (i == asz) {
      kill();
    }
    for (int k = i + 1; k < asz; k++) {
      if ((bv[j].first ^ av[i].first) & lsb(av[k].first)) {
        av[i].first ^= av[k].first;
        A.act(i, k);
      }
    }
    if (av[i].first != bv[j].first) {
      kill();
    }
    A.swap(i, j);
    i++;
  }
  for (int i = bsz; i < asz; i++) {
    A.act(i, i);
  }
}
int main() {
  scanf("%d", &N);
  A.compute();
  B.compute();
  transfer();
  vector<pair<int, int> > ans = A.ops;
  ans.insert(ans.end(), B.ops.rbegin(), B.ops.rend());
  printf("%lu\n", ans.size());
  for (pair<int, int> p : ans) {
    printf("%d %d\n", p.first + 1, p.second + 1);
  }
}